Chapter 8, Problem 13CT

Find the area of the shaded region enclosed in a semicircle of diameter $8$ centimetres. The length of the chord $AB$ is $6$ centimetres.

[Hint: Triangle $ABC$ is a right triangle.]

To determine

The area of the shaded region enclosed in a semicircle of diameter $8$ centimeters and the length of the chord $AB$ is $6$ centimeters.

Answer to Problem 13CT

Solution:

The area of the shaded region is $9.2582c{m}^2$

Explanation of Solution

Given information:

A shaded region is as shown in the figure below:

A semicircle of diameter $8$ centimeters and

The length of the chord $AB$ is $6$ centimeters.

$△ABC$ is a right triangle

Explanation:

To find the area of the shaded region, subtract the area of the triangle $△ABC$ from the area of the semicircle.

Area of semicircle is $A=\frac{1}{2}\pi r^2$

Here, diameter of the semicircle is $8$ centimeters

$\Rightarrow r=\frac{D}{2}=\frac{8}{2}=4$

By substituting value of $r=4$ in the area of semicircle,

$\Rightarrow A=\frac{1}{2}\pi \cdot 4^2$

$\Rightarrow A=8\pi$

$\Rightarrow A=25.1327$

Now, $△ABC$, $\angle B$ is the right angle

Thus, by using Pythagorean Theorem $c^2=a^2+b^2$

$\Rightarrow 8^2=6^2+{\left(BC\right)}^2$

$\Rightarrow 64=36+{\left(BC\right)}^2$

$\Rightarrow {\left(BC\right)}^2=64-36$

$\Rightarrow {\left(BC\right)}^2=28$

$\Rightarrow \left(BC\right)=2\sqrt{7}$

Thus, length of $BC=2\sqrt{7}$

Area of triangle $A\left(\Delta ABC\right)=\frac{1}{2}\cdot 6\cdot \left(AB\right)$

$\Rightarrow A\left(\Delta ABC\right)=\frac{1}{2}\cdot 6\cdot 2\sqrt{7}$

$\Rightarrow A\left(\Delta ABC\right)=6\sqrt{7}$

$\Rightarrow A\left(\Delta ABC\right)=15.8745$

Therefore, area of the shaded region $=25.1327-15.8745$

$=9.2582$

Thus, the area of the shaded region is $9.2582c{m}^2$.